Based on the book Computer Vision a Modern Approach page 425, I attempted to use eigenvectors for image segmentation.
http://dl.dropbox.com/u/1570604/tmp/comp-vis-modern-segment.pdf
The author mentions that image pixel affinites can be captured in matrix A. Then we can maximize w^T A w product where w's are weights. After some algebra one obtains Aw = \lambda w, finding w is like finding eigenvectors. Then finding the best cluster is finding the eigenvalue with largest eigenvector, the values inside that eigenvector are cluster membership values. I wrote this code
import matplotlib.pyplot as plt
import numpy as np
Img = plt.imread("twoObj.jpg")
(n,dummy) = Img.shape
Img2 = Img.flatten()
(nn,) = Img2.shape
A = np.zeros((nn,nn))
for i in range(nn):
for j in range(nn):
N=Img2[i]-Img2[j];
A[i,j]=np.exp(-(N**2))
V,D = np.linalg.eig(A)
V = np.real(V)
a = np.real(D[1])
threshold = 1e-10 # filter
a = np.reshape(a, (n,n))
Img[a<threshold] = 255
plt.imshow(Img)
plt.show()
The image
Best result I could get from this is below. I have a feeling the results can be better.
The eigenvalues are sorted from largest to smallest in Numpy, I tried the first one, that did not work, then I tried the second one for the results seen below. Threshold value was chosen by trial and error. Any ideas on how this algorithm can be improved?
I've just tried the algorithm in Mathematica, it works fine on your image, so there must be a subtle bug in your code.
This part:
looks strange: all linear algebra packages I know return the eigenvalues/eigenvectors sorted, so you'd just take the first eigenvector in the list. That's the one that corresponds to the highest eigenvalue. Try plotting the eigenvalues list to confirm that.
Also, where did you get the fixed threshold from? Have you tried normalizing the image to display it?
For what it's worth, the results I'm getting for the first 3 eigenvectors are:
This is the Mathematica code I use: